关于python中的链表

Question

I need to create a function to remove a node from a linked list from a value given manually inside this function but I don't know exactly how to do that. My algorithm is like this:

noRaiz = None

def novoNo(valor):
    return {
        "valor": valor,
        "proximo": None
    }
def remove(valor):
global noRaiz
if noRaiz is None:
    return
noAtual = noRaiz
if noRaiz["valor"] == valor:
    noRaiz = noRaiz["proximo"]


def imprimir():
    noAtual = noRaiz
    while noAtual is not None:
        print(noAtual["valor"])
        noAtual = noAtual["proximo"]


noRaiz = novoNo(54)
no2 = novoNo(26)
no3 = novoNo(93)
no4 = novoNo(17)
no5 = novoNo(77)
no6 = novoNo(31)

noRaiz["proximo"] = no2
no2["proximo"] = no3
no3["proximo"] = no4
no4["proximo"] = no5
no5["proximo"] = no6
no6["proximo"] = None


imprimir()

noRaiz is the first node on the list, noAtual is the current node in the list at print time when running the code and imprimir() it's the function to print the nodes. What would the function look like if I wanted to remove node 54 or 17?

Answer 1

The trick is to move through the linked list and look one node ahead for a match with the value. When there is a match, you should alter the link of your current node so that it skips that next node having the value.

Here is the function:

def remove(valor):
    global noRaiz
    if noRaiz is None:  # Nothing to do
        return
    noAtual = noRaiz
    if noRaiz["valor"] == valor:  # Special case
        noRaiz = noRaiz["proximo"]
        return
    while noAtual["proximo"] is not None:
        if noAtual["proximo"]["valor"] == valor:
            noAtual["proximo"] = noAtual["proximo"]["proximo"]
            break
        noAtual = noAtual["proximo"]

Remark: consider implementing your linked list with a Node class instead of a global dictionary.